Number systems
- Place values
- binary 2k, octal 8k, hex 16k
- To decimal
- Σ dk·bk
- From decimal
- divide by the base, read the remainders bottom-up
- Hex ↔ binary
- one hex digit = 4 bits
- Two’s complement
- invert the bits and add 1
Formula sheets, the key method steps for every topic and 20 practice problems, with full solutions at the back.
Set the bits so they store −25 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
A system must give every one of 2677 items its own binary code of the same length. What is the smallest number of bits b needed?
In how many of the 8 rows of its truth table is (p ∨ (r ∨ q)) → ((p → q) ∨ ¬r) true? Call this number N.
The premises (q ∧ r) → p; q; ¬p are given. Which conclusion follows from them? Click it.
Use the Karnaugh map to find the minimal sum-of-products (SOP) form of F(A, B, C) = Σm(1, 2, 4, 5, 6). How many product terms N does it have?
A 4-bit ring counter Q₃Q₂Q₁Q₀ starts at 1100. At every clock edge the bits shift one place to the right (Q₃ → Q₂ → Q₁ → Q₀) and Q₃ receives Q₀. Give the state Q = Q₃Q₂Q₁Q₀ as a decimal number after 4 clock edges.
Solve 65x ≡ 88 (mod 97). Give the solution x with 0 < x < 97.
Encrypt the message m = 39 with the RSA public key (n, e) = (187, 17). Find the ciphertext c.
What is the time complexity of the standard dynamic programming algorithm for the longest common subsequence of strings of lengths m and n?
A problem is NP-complete if it is:
Why is traversing an array often much faster in practice than traversing a linked list of the same length?
A dynamic array doubles when full. Why do many implementations shrink it only when it is one quarter full, not half full?
A tree has 2 vertices of degree 3, 2 vertices of degree 4, 4 vertices of degree 5, and all its other vertices are leaves (degree 1). How many leaves L does it have?
How many closed walks of length 3 start and end at E? A walk may repeat vertices and edges. Call the number N.
In relational algebra, the projection operator (π) corresponds to:
Which SQL feature is best suited to finding every employee below a manager in an organisation tree of unknown depth?
How many usable host addresses does a /24 IPv4 network have?
Head-of-line blocking in HTTP/2 over TCP occurs because:
The network 192.168.243.0/24 is divided into /25 subnets, numbered 1, 2, 3, … from the lowest address (subnet 1 starts at .0). What is the fourth octet x of the broadcast address of subnet 2?
The network 192.168.73.0/24 is split with VLSM for these LANs: A: 34 hosts, B: 9 hosts, C: 6 hosts, D: 19 hosts, E: 12 hosts. Subnets are allocated from .0 upwards, the LAN with the most hosts first, each with the smallest subnet that fits. What is the fourth octet x of the network address of LAN D?
Answer: U = 231
Setting bits: each bit is worth a power of two (128, 64, …, 1). One hex digit is exactly four bits, and in 8-bit two's complement the leftmost bit is worth −128. Bitwise operations work on each bit position separately.
Answer: b = 12 bit
Positional number systems: each digit is worth its value times a power of the base. In binary the places are worth 1, 2, 4, 8, 16, …; in hexadecimal 1, 16, 256, …, with A–F standing for 10–15. In 8-bit two's complement the leftmost bit is worth −128 instead of +128, which is how negative numbers are stored.
Answer: N = 7
Truth tables: a formula with n different variables has 2ⁿ rows, one for every combination of truth values (1 = true, 0 = false). Work from the inside out and give every subformula its own column. ¬a flips the value; a ∧ b is 1 only when both are 1; a ∨ b is 0 only when both are 0; a → b is 0 only when a = 1 and b = 0; a ↔ b is 1 when both sides are equal.
Answer: ¬r
Quantifiers and arguments: ¬∀x P(x) ≡ ∃x ¬P(x) and ¬∃x P(x) ≡ ∀x ¬P(x). Move a negation inwards, swap every quantifier it passes, and negate the inner formula with the propositional laws, such as ¬(a → b) ≡ a ∧ ¬b. An argument is valid when the conclusion is 1 in every row in which all premises are 1. A single row with true premises and a false conclusion is a counterexample and shows the argument is invalid.
Answer: N = 3
Karnaugh maps: the cells are ordered in Gray code (00, 01, 11, 10), so neighbouring cells, including the ones across the edges, differ in exactly one variable. Group the 1s in rectangles of 1, 2, 4, 8 or 16 cells, as large as possible and as few as possible, until every 1 is covered. Each group is one product term: the variables that stay constant inside it. A group of 2ᵏ cells loses k variables. Don't-cares (X) may be used to make groups bigger but never have to be covered. Grouping the 0s instead gives the minimal POS form.
Every group is a rectangle of 2ᵏ cells; edge cells are neighbours of the cells on the opposite edge.
Another grouping of the same size exists; it has the same number of terms and literals.
Answer: Q = 12
Flip-flops and counters: a flip-flop changes only at the active clock edge. D copies its input (Q⁺ = D); T toggles when T = 1; JK holds for 00, resets for 01, sets for 10 and toggles for 11. An n-bit binary counter has 2ⁿ states and counts modulo 2ⁿ; a mod-N counter goes 0, 1, …, N − 1 and back to 0, and needs the smallest n with 2ⁿ ≥ N flip-flops. Each stage of a ripple counter halves the frequency, so the last output of a mod-N counter runs at fclk/N. A Johnson counter of n flip-flops has 2n states, a ring counter n.
Answer: x = 70
Inverses and congruences: a has an inverse modulo n exactly when gcd(a, n) = 1. The extended Euclidean algorithm writes 1 = s·n + t·a, and then a⁻¹ ≡ t (mod n). A congruence a·x ≡ b (mod n) with gcd(a, n) = 1 has the solution x ≡ a⁻¹·b. Two congruences x ≡ r₁ (mod m₁) and x ≡ r₂ (mod m₂) with coprime moduli have one solution modulo m₁·m₂ (Chinese remainder theorem). Euler’s φ(n) counts the numbers from 1 to n that are coprime to n: φ(n) = n·∏(1 − 1/p).
Answer: c = 107
Modular powers and RSA: compute aᵏ mod n by repeated squaring: write k in binary, square, reduce after every step and multiply the powers whose bit is 1. Fermat: aᵖ⁻¹ ≡ 1 (mod p) for a prime p not dividing a, so the exponent can be reduced modulo p − 1. RSA: choose primes p and q, n = p·q and φ(n) = (p − 1)(q − 1); the public exponent e is coprime to φ(n), and the private exponent is d = e⁻¹ mod φ(n). Encrypt with c = mᵉ mod n and decrypt with m = cᵈ mod n.
Answer: A · O(mn)
The table has (m + 1)(n + 1) cells and each is computed in O(1) from its neighbours.
Answer: C · In NP and every NP problem reduces to it in polynomial time
NP-complete problems are the hardest in NP; a polynomial algorithm for one would give one for all of NP.
Answer: D · Array elements are contiguous, so they make good use of the CPU cache
Contiguous memory gives good locality of reference, while linked nodes may be scattered and cause many cache misses.
Answer: C · To avoid repeated resizing when pushes and pops alternate near the boundary
Shrinking at one half could trigger a resize on every operation in such a sequence; the gap between 1/4 and full keeps the amortised cost O(1).
Answer: L = 20
Euler, Hamilton and trees: a connected graph has an Euler circuit (every edge once, back to the start) when every vertex has even degree, and an Euler path that is not a circuit when exactly two vertices have odd degree; the path then runs from one odd vertex to the other. A Hamilton circuit visits every vertex once; Kn has (n − 1)!/2 of them. A tree on n vertices has n − 1 edges, and a connected graph with m edges keeps a spanning tree after removing m − n + 1 edges.
Shortcut: L = 2 + Σ (d − 2) over the vertices with degree d ≥ 3.
Answer: N = 2
Degrees: the degree of a vertex is the number of edges that meet it. Every edge has two ends, so the degrees add up to twice the number of edges: Σ deg(v) = 2|E| (the handshake lemma). The number of vertices with odd degree is therefore even. Kn has n(n − 1)/2 edges, a k-regular graph on n vertices has nk/2, and a tree on n vertices has n − 1 edges.
Answer: C · Choosing which columns appear in the result
Projection keeps chosen attributes (the SELECT list), while selection σ keeps rows that satisfy a condition (the WHERE clause).
Answer: A · A recursive common table expression (WITH RECURSIVE)
A recursive CTE starts with an anchor query and repeatedly joins the previous result to the table, so it handles hierarchies of any depth.
Answer: B · 254
A /24 has 8 host bits, so 2⁸ = 256 addresses. The network and broadcast addresses are reserved, leaving 254 for hosts.
Answer: C · A lost TCP segment delays all multiplexed streams until it is retransmitted
TCP delivers bytes in order, so one loss stalls every stream on the connection. QUIC (HTTP/3) handles streams independently over UDP.
Answer: x = 255
IPv4 subnetting: an address has 32 bits. The prefix /p says how many leading bits identify the network; the remaining 32 − p bits number the hosts. A subnet has 2³²⁻ᵖ addresses, but the first (network address) and last (broadcast address) cannot be given to hosts, so 2³²⁻ᵖ − 2 hosts fit. Exceptions: a /31 (point-to-point link, RFC 3021) uses both of its 2 addresses for hosts, and a /32 is one single address (a single host or a loopback).
Answer: x = 64
VLSM: each LAN gets the smallest subnet with 2ʰ − 2 ≥ its host count, i.e. prefix 32 − h. Allocating the largest subnets first keeps every block aligned on a multiple of its own size, so the network addresses follow by adding up the block sizes.
LAN D: 192.168.73.64/27.
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